lecture1.tex

%% -*- coding:utf-8 -*-
\chapter{Generalities on algebraic extensions}
We introduce the basic notions such as a field extension, algebraic
element, minimal polynomial, finite extension, and study their very
basic properties such as the multiplicativity of degree in towers. 

\section{Field extensions: examples}

\subsection{K-algebra}
\begin{definition}[K-algebra]
  \label{def:kalgebra}
  Let $K$ is a \mynameref{def:ring} and $A$ is a
  \mynameref{def:module} over $K$ 
  \footnote{
    The definition in the lectures is more common and assumes that $K$
    is a ring but later $K$ will be mostly a \mynameref{def:field}. Therefore
    $A$ has to be a \mynameref{def:vectorspace} over $K$ in the case. 
  }
  equipped with 
  an additional binary operation $A \times A \rightarrow A$ that we
  denote as $\cdot$ here. The $A$ is an algebra over $K$ if the
  following identities hold $\forall x,y,z \in A$ and for every
  elements (often called as scalar) $a, b \in K$
  \begin{itemize}
  \item Right distributivity:
    $(x + y) \cdot z = x \cdot z + y \cdot z$
  \item Left distributivity:
    $z \cdot (x + y) = z \cdot x + z \cdot y$
  \item Compatibility with scalars:
    $(ax) \cdot (by) = (ab) (x \cdot y)$
  \end{itemize}
\end{definition}

\begin{example}[Field of complex numbers $\mathbb{C}$]
  The field of complex numbers $\mathbb{C}$ can be considered as a
  K-algebra over the field of real numbers $\mathbb{R}$.
  \label{ex:complexnumbers}
\end{example}

\subsection{Field extension}

\begin{definition}[Field extension]
  Let $K$ and $L$ are fields.
  $L$ is an extension of $K$ if $L \supset K$
  \label{def:fextension1}
\end{definition}
and another definition
\begin{definition}[Field extension]
  Let $K$ is a field then
  $L$ is an extension of $K$ if $L$ is a K-algebra
  \footnote{
    $L$ in the definition is not the same object with $L$ from
    definition \ref{def:fextension1}. Because $L$ in the definition is
    a $K$-algebra i.e. a ring but $L$ in the definition
    \ref{def:fextension1} is a field. 
  }
  \label{def:fextension2}
\end{definition}
Why the 2 definitions are equivalent?

\begin{lemma}[K-algebra and homomorphism]
  Given a K-algebra is the same as
  having \mynameref{def:homomorphism} $f: K \rightarrow A$ of rings.
  \begin{proof}
    Really if I have a K-algebra I can define the
    \mynameref{def:homomorphism} $f(k) = k \cdot 1_A$, where $1_A$ is an
    identity element of $A$. Thus $k \cdot 1_A \in A$.

    And conversely if I have the \mynameref{def:homomorphism}
    $f: K \rightarrow A$
    I can define the K-algebra structure by setting
    $k a = f(k) a$ because $f(k), a \in A$ and there is a multiplication
    defined on $A$. As result I have a rule for multiplication a vector
    ($f(k) \in A$) on a vector ($a \in A$) that is required for
    \mynameref{def:kalgebra} setup.
  \end{proof}
  \label{lem:lec1_homkalgebra}
\end{lemma}

\begin{lemma}[About homomorphism of fields]
  Any \mynameref{def:fieldhomomorphism} is
  \mynameref{def:injection}.
  \footnote{
    But the statement is not valid for groups. In that case the
    homomorphism is injection if and only if the 
    kernel is trivial and consists of only one element - identity.
    See theorem \ref{thm:grouphomomorphsim}.
  }
  \label{lem:lec1_homomorphism_is_injection}
  \begin{proof}
    Lets proof by contradiction.
    Really if $f(x) = f(y)$ and $x \ne y$ then
    \begin{eqnarray}
      f(x) - f(y) = 0_A,
      \nonumber \\
      f(x - y) = 0_A,
      \nonumber \\
      f(x - y) f(\left(x - y\right)^{-1}) =
      f\left(\frac{x - y}{x - y}\right) = f(1_K) = 1_A = 0_A
      \nonumber
    \end{eqnarray}
    that is impossible.
  \end{proof}  
\end{lemma}

There are some comments on the results. We have got that a
\mynameref{def:homomorphism} can be set between field $K$ and its
K-algebra.
The \mynameref{def:homomorphism} is \mynameref{def:injection}
therefore we can allocate a sub-field $A' \subset A$ for that we will
have the \mynameref{def:homomorphism} is a \mynameref{def:surjection} and
therefore we have an \mynameref{def:isomorphism} between original field $K$
and a sub-field $A'$. This means that we can say that the original
field $K$ is a sub-field for the K-algebra.

\begin{example}[Field extensions]
  $\mathbb{C}$ is a field extension for $\mathbb{R}$.
  $\mathbb{R}$ is a field extension for $\mathbb{Q}$
  \label{ex:fieldextension}
\end{example}

\begin{example}[$K$-algebra is not a field]
  In the example
  \footnote{the example was not present in the lectures}
  I will show
  that $K$ algebra is not a field. Consider
  $K = \mathbb{R}$. \mynameref{def:vectorspace} $A = \mathbb{R}^2$ i.e.
  $A$ consists of vectors of the following form
  \[
  x = \left(
  \begin{array}{c}
    x_1 \\
    x_2
  \end{array}
  \right),
  \]
  where $x_1, x_2 \in \mathbb{R}$.
  I will define the multiplication for $L$ (our $K$ algebra) as
  follows
  \[
  \left(
  \begin{array}{c}
    x_1 \\
    x_2
  \end{array}
  \right)
  \cdot
  \left(
  \begin{array}{c}
    y_1 \\
    y_2
  \end{array}
  \right) =
  \left(
  \begin{array}{c}
    x_1 \cdot y_1 \\
    x_2 \cdot y_2
  \end{array}
  \right)
  \]

  It can be seen that all requirements of \mynameref{def:kalgebra} are
  satisfied
  \begin{eqnarray}
    (x + y) \cdot z =
    \left(
    \left(
    \begin{array}{c}
      x_1 \\
      x_2
    \end{array}
    \right)
    +
    \left(
    \begin{array}{c}
      y_1 \\
      y_2
    \end{array}
    \right)
    \right)
    \left(
    \begin{array}{c}
      z_1 \\
      z_2
    \end{array}
    \right) =
    \nonumber \\
    =
    \left(
    \begin{array}{c}
      (x_1 + y_1)z_1 \\
      (x_2 + y_2)z_2
    \end{array}
    \right) =
    \left(
    \begin{array}{c}
      x_1 z_1 + y_1 z_1 \\
      x_2 z_2 + y_2 z_2
    \end{array}
    \right) =
    x \cdot z + y \cdot z
    \nonumber \\
    z \cdot (x + y) = z \cdot x + z \cdot y
    \nonumber \\
    (ax) \cdot (by) =
    \left(
    \begin{array}{c}
      a x_1 \\
      a x_2
    \end{array}
    \right)
    \left(
    \begin{array}{c}
      b y_1 \\
      b y_2
    \end{array}
    \right) =
    \left(
    \begin{array}{c}
      ab x_1 y_1 \\
      ab x_2 y_2
    \end{array}
    \right)
    (ab) (x \cdot y)
    \nonumber
  \end{eqnarray}
  The multiplication identity element of $L$ is the following
  \[
  1_L = 
  \left(
  \begin{array}{c}
    1 \\
    1
  \end{array}
  \right)
  \]
  The zero is the standard one from vector space
  \[
  0_L = 
  \left(
  \begin{array}{c}
    0 \\
    0
  \end{array}
  \right)
  \]

  We can see that
  \[
  \left(
  \begin{array}{c}
    1 \\
    0
  \end{array}
  \right)
  \left(
  \begin{array}{c}
    0 \\
    1
  \end{array}
  \right) = 0_L
  \]
  i.e. we have 2 divisor of zero which are not zero itself. The
  elements do not have invert ones and as result the $L$ is
  not a field.

  From other side if we define $L' \subset L$ as follows
  $L' = \left\{
  \left(
  \begin{array}{c}
    r \\
    r
  \end{array}
  \right)
  \right\}$, where $r \in \mathbb{R}$,
  then we will have that $L'$ is a field and $L' \cong \mathbb{R}$.
  \label{ex:kalgebranotfield}
\end{example}

\begin{definition}[Characteristic]
  \label{def:fieldcharacteristic}
  If $L$ is a field there are 2 possibilities
  \begin{enumerate}
  \item $1 + 1 + \dots \ne 0$. In this case
    $\mathbb{Z} \subset L$ but $\mathbb{Z}$ is not a field therefore $L$
    is an extension of $\mathbb{Q}$. In the case $char L = 0$
  \item $1 + 1 + \dots + 1 = \sum_{i = 1}^m 1 = 0$ for some $m \in
    \mathbb{Z}$. The first time when it happens is for a prime number
    i.e. minimal $m$ with the property is prime. In this case $char L
    = p$, where $p = \min m$  - the minimal $m$ (prime) with the
    property. In this 
    case $\mathbb{Z}/p\mathbb{Z} \subset L$. The
    $\mathbb{Z}/p\mathbb{Z}$ is a field denoted by $\mathbb{F}_p$. The
    $L$ is an extension of $\mathbb{F}_p$.
  \end{enumerate}
  No other possibilities exist. The $\mathbb{Q}$ and $\mathbb{F}_p$ are
  the prime fields. Any field is an extension of one of those.
\end{definition}

\begin{definition}[Prime field]
  \label{def:primefield}
  If $L$ is a field $\mathbb{Q} \subset L$, i.e. $char L = 0$ then
  $\mathbb{Q}$ is the prime field for $L$. Otherwise, if $char L = p > 0$
  then $\mathbb{F}_p$ is the prime field.
\end{definition}

\begin{claim}
\label{claim:lec1_sec14}
Let $K\left[X\right]$ \mynameref{def:ring} of polynomials.
The $P \in K\left[X\right]$ is an
\mynameref{def:irreducible}. $\left(P\right)$ is an 
\mynameref{def:ideal} formed by the polynomial.
\footnote{
  I.e.
  \(
  \left(P\right) = \left\{ Q = G P \right\} 
  \)
  where $G \in K\left[X\right]$
}
The set of residues by
the polynomial forms a field that denoted by
$K\left[X\right]/\left(P\right)$.
\begin{proof}
How we can see it?
If $Q \in K\left[X\right]$ is a polynomial that $Q \notin
\left(P\right)$ when $Q$ is prime to $P$.
\footnote{
  As soon as $P$ is irreducible in $K\left[X\right]$ then there is only
  one possibility for $Q$ and $P$ to have common divisors: if $Q = G P$
  where $G \in K\left[X\right]$ but this is in contradiction with
  $Q \notin \left(P\right)$
}
Then with
\mynameref{lem:bezout} we can get $\exists A, B \in K\left[X\right]$
such that
\[
A P + B Q = 1,
\]
or
\[
B Q \equiv 1 \mod P,
\]
thus $B$ is $Q^{-1}$ in $K\left[X\right]/\left(P\right)$.
\end{proof}
\end{claim}

\begin{claim}
  \footnote{
    The claim is not a part of the lectures but it's very usefully in
    future lectures.
  }
  Let $K$ is a field and $a \in K$ then $K\left[X\right]/\left(X -
  a\right)$ is also a field and there exists an
  \mynameref{def:isomorphism} between the field and $K$ i.e.
  \begin{equation}
    K\left[X\right]/\left(X - a\right) \cong K
    \label{eq:lec1_ex1_1}
  \end{equation}
  \label{claim:lec1_fieldquotionisomorphism}
  \begin{proof}
    The $K\left[X\right]/\left(X - a\right)$ is a field just because
    $X-a$ is
    \mynameref{def:irreducible} (see example \ref{ex:irreducible} and
    claim \ref{claim:lec1_sec14}).

    For the proof the main statement (\ref{eq:lec1_ex1_1}) lets consider a
    polynomial $P \in  K\left[X\right]$ and 
    define the following \mynameref{def:fieldhomomorphism}:
    \begin{equation}
      \phi: K\left[X\right]/\left(X - a\right)
      \xrightarrow[P(X) \to P(a)]{} K
      \label{eq:lec1_ex1_2}
    \end{equation}

    The $\phi$ defined by (\ref{eq:lec1_ex1_2}) is
    \mynameref{def:fieldhomomorphism}. For the proof of the claim lets take
    $P_1, P_2 \in K\left[X\right]/\left(X - a\right)$. Clear that
    \[
    \phi\left(P_1 + P_2\right) = P_1\left(a\right) + P_2\left(a\right)
    = \phi\left(P_1\right) + \phi\left(P_1\right).
    \]
    The same holds with the multiplication:
    \[
    \phi\left(P_1 \cdot P_2\right) = P_1\left(a\right) \cdot P_2\left(a\right)
    = \phi\left(P_1\right) \cdot \phi\left(P_1\right).
    \]
    %% For the inverse element image lets consider $P \in
    %% K\left[X\right]/\left(X - a\right)$ such that $P \ne 0$ and as
    %% result $\phi\left(P\right) = P\left(a\right) \ne 0$.
    %% There exists $P^{-1} \in K\left[X\right]/\left(X - a\right)$ as
    %% soon as $K\left[X\right]/\left(X - a\right)$ is a field. Therefore
    %% we have
    %% \begin{eqnarray}
    %%   \phi(1) = \phi\left(P P^{-1}\right) =
    %%   \phi\left(P\right)\phi\left(P^{-1}\right) =
    %%   \nonumber \\
    %%   = P\left(a\right) P^{-1}\left(a\right) = 1,
    %%   \nonumber
    %% \end{eqnarray}
    %% i.e.
    %% \[
    %% \phi\left(P^{-1}\right) = \phi\left(P\right)^{-1}.
    %% \]
    That completes the proof of the \mynameref{def:fieldhomomorphism}.
    %% Division is more complex but also can be
    %% shown: if $P_2 \ne 0$ when there exists
    %% $P_2^{-1} \in K\left[X\right]/\left(X - a\right)$ as soon as
    %% $K\left[X\right]/\left(X - a\right)$ is the field then
    %% with $\phi\left(P_2^{-1}\right) = P_2^{-1}\left(a\right) =
    %% \frac{1}{\phi\left(P_2\right)}$ one can get
    %% \[
    %% \phi\left(\frac{P_1}{P_2}\right) =
    %% \phi\left(P_1 P_2^{-1}\right) =
    %% \phi\left(P_1\right) \phi\left(P_2^{-1}\right) =
    %% \frac{\phi\left(P_1\right)}{\phi\left(P_2\right)}
    %% \]

    %% We have $\ker{\phi} = \left(X - a\right)$ because for any polynomial
    %% $P$ that is in the ideal $\left(X - a\right)$ has $P(a) = 0$ i.e. in
    %% the kernel of $\phi$.

    Lemma \mynameref{lem:lec1_homomorphism_is_injection} says that any
    \mynameref{def:fieldhomomorphism} is \mynameref{def:injection},
    i.e. $\phi$ is an injection.
    
    Next we should show that $\phi$ is \mynameref{def:surjection} it's
    easy because $\forall k \in K$ we can consider constant polynomial
    $P = k$ from $K\left[X\right]$. For the polynomial we will have
    $\phi\left(k\right) = k$.

    As result $\phi$ is \mynameref{def:injection} as well as
    \mynameref{def:surjection} i.e. $\phi$ is
    \mynameref{def:bijection}. With the fact that $\phi$ is
    \mynameref{def:homomorphism} it will gives us that $\phi$ is
    \mynameref{def:isomorphism}. 
    %% Now (\ref{eq:lec1_ex1_1}) follows from the
    %% \mynameref{thm:firstisomorphism}. 
  \end{proof}
\end{claim}

\section{Algebraic elements. Minimal polynomial}

\subsection{$K\left[X\right]/\left(P\right)$ field}

Alternative proof of claim \ref{claim:lec1_sec14} i.e. the fact that
$K\left[X\right]/\left(P\right)$ is the 
\mynameref{def:field} as follows.
The $\left(P\right)$ is a \mynameref{def:maxideal}
\footnote{
  To prove that $\left(P\right)$ is a \mynameref{def:maxideal} we have
  to use \mynameref{lem:bezout}.
}
but a quotient by a  \mynameref{def:maxideal} is a \mynameref{def:field}
(see \mynameref{thm:maxideal}).

We also can say that $K\left[X\right]/\left(P\right)$ is an extension
of $K$ because it's 
\mynameref{def:kalgebra}.

\begin{example}[$K = \mathbb{F}_2/\left(X^2+X+1\right)$]
  Lets consider the following field
  $\mathbb{F}_2 = \mathbb{Z}/2\mathbb{Z} = \{0,1\}$ in the
  field polynomial $X^2+X+1$ is irreducible. It's very easy to verify
  it because $\mathbb{F}_2$ has only 2 elements that can be (possible)
  a root:
  \[
  0^2+0+1 = 1\ne 0
  \]
  and
  \[
  1^2+1+1 = 1\ne 0
  \]
  
  The polynomial has the following
  residues: $\bar{X} = X + \left(X^2+X+1\right)$ and
  $\overline{X + 1} = X + 1 + \left(X^2+X+1\right)$. Thus the field
  $\mathbb{F}_2/\left(X^2+X+1\right)$ consists of 4 elements:
  $\{0, 1, \bar{X}, \overline{X+1}\}$.

  It's easy to see that the third element ($\bar{X}$) is a root of
  $P(X) = X^2+X+1$:
  \[
  \bar{X}^2 + \bar{X} + 1 =
  P(X) + \left(P(X)\right) = \left(P(X)\right) \equiv 0 \mod P.
  \]
  
  \[
  \bar{X}^2 + \bar{X} + 1 = \bar{0},
  \]
  therefore 
  \[
  \bar{X}^2 = - \bar{X} - 1 = \bar{X} + 1 = \overline{X+1}.
  \]
  This is because we are in field $\mathbb{F}_2$ where
  \[
  2 \left(X + 1\right) \mod 2 = 0 
  \]
  and thus
  \[
  - \bar{X} - 1 = \bar{X} + 1 = \overline{X + 1}
  \]
  
  Also
  \[
  \overline{X+1}^2 = \bar{X},
  \]
  and they are inverse each other
  \[
  \overline{X+1} \bar{X} = 1,
  \]

  So this is the structure of a field of four elements.  The
  cardinality of $K=\mathbb{F}_2/\left(X^2+X+1\right)$ is 4,  one
  writes then $K=F_4$. Well, this might be 
  strange at the first sight, because we only know that $K$ has four
  elements and if you write $F_4$  you somehow mean that there is only
  one field of four elements. Well, it is true, there is only one
  field of four elements. In fact, all finite fields of the same
  cardinality are isomorphic, and we will see it very shortly
  (see theorem \ref{thm:lec3_1}).  
  \label{ex:F2overP}
\end{example}

Arguments for the \autoref{ex:F2overP} can also be found in the
following remark (not part of the lectures)
\begin{remark}[About quotient field]
Consider an arbitrary field $L$ ($L=\mathbb{F}_2$ at the
\autoref{ex:F2overP}) and $P\left(X\right) = X^n + a_{n-1}X^{n-1} +
a_{n-2} X^{n-2} + \dots + a_0 \in L\left[X\right]$ is an
irreducible polynomial over $L$. Let $\alpha$ is a root of the
polynomial then 
\[
\alpha^n + a_{n-1} \alpha^{n-1} + \dots + a_0 = 0,
\]
or
\[
\alpha^n = - a_{n-1} \alpha^{n-1} - \dots - a_0.
\]
Therefore $\forall l \ge n, \exists l_i \in L$ such that 
\[
\alpha^k = \sum_{i=0}^{n-1}l_i \alpha^{i}
\]
and as result $\forall x \in L/\left(P\right)$ exists $x_i \in L$
such that 
\[
x = \sum_{i=0}^{n-1}x_i \alpha^{i}.
\]
We can say that we have the following basis: $\{\{L\}, \alpha, \dots,
\alpha^{n-1}\}$ where $\{L\}$ - L basis. For \autoref{ex:F2overP} we
have
$\{L\}= \{\mathbb{F}_2\}= \{0,1\}$, $\alpha = \bar{X}$ and as result
$K = \mathbb{F}_2/\left(X^2+X+1\right)$ will have the following basis
$\{0,1,\bar{X}\}$.
\end{remark}

\subsection{Algebraic elements}

\begin{definition}[Algebraic element]
  Let $K \subset L$ and $\alpha \in L$. $\alpha$ is an algebraic
  element if $\exists P \in K\left[X\right]$ such that
  $P\left(\alpha\right) = 0$. Otherwise the $\alpha$ is called
  transcendental.
  \label{def:algebraicelement}
\end{definition}

\subsection{Minimal polynomial}

\begin{lemma}[About minimal polynomial existence]
  If $\alpha$ is \mynameref{def:algebraicelement} then
  $\exists!$ unitary polynomial $P$ of minimal degree such that
  $P\left(\alpha\right) = 0$. It is irreducible. $\forall Q$ such that
  $Q\left(\alpha\right) = 0$ is divisible by $P$
  \footnote{
    see also \mynameref{thm:irreduciblediv}
  }
  \begin{proof}
    We know that $K\left[X\right]$ is a \mynameref{def:pid} and a
    polynomial $Q\left(\alpha\right) = 0$ forms an
    \mynameref{def:ideal}: $I \left\{Q \in K\left[X\right] \mid
    Q\left(\alpha\right) = 0 \right\}$, so the ideal is generated by
    one element: $I = \left(P\right)$. This is an unique (up to
    constant) polynomial minimal degree in $I$.
    
    Lets prove that $P$ is irreducible. If $P$ is not irreducible then
    $\exists Q,R \in I$ such that $P = Q 
    R$, $Q(\alpha) = 0$ or $R(\alpha) = 0$ and
    $deg R,Q < deg P$ that is in contradiction with the definition
    that $P$ is a polynomial of minimal degree.    
  \end{proof}
  \label{lem:minpolynomial}
\end{lemma}

\begin{definition}[Minimal polynomial]
  If $\alpha$ is \mynameref{def:algebraicelement} then
  the unitary polynomial $P$ of minimal degree such that
  $P\left(\alpha\right) = 0$ is called minimal polynomial and denoted
  by  $P_{min}\left(\alpha, K\right)$.
  \label{def:minpolynomial}
\end{definition}

\begin{gapexample}[Minimal polynomial]
  Consider the following minimal polynomial
  $P_{min}\left(\alpha, \mathbb{Q}\right)$, where $\alpha =
  \sqrt{2}$. In GAP \cite{url:gap} we have
\begin{verbatim}
  gap> x:=Indeterminate(Rationals,"x");;
  gap> alpha:=Sqrt(2);;
  gap> MinimalPolynomial(Rationals, alpha);
  x^2-2
\end{verbatim}
I.e. $P_{min}\left(\alpha, \mathbb{Q}\right) = X^2 -2$.
\label{ex:minpolynomial}
\end{gapexample}

\section{Algebraic elements. Algebraic extensions}

\begin{definition}
  Let $K \subset L$, $\alpha \in L$. The smallest sub-field contained
  $K$ and $\alpha$ denoted by $K\left(\alpha\right)$. The smallest
  sub-ring (or \mynameref{def:kalgebra}) contained $K$ and $\alpha$ denoted by $K\left[\alpha\right]$.
\end{definition}

As soon as $K\left[\alpha\right]$ is a \mynameref{def:kalgebra} it is a
\mynameref{def:vectorspace} over $K$ generated by
\[
1, \alpha, \alpha^2, \dots,
\alpha^n, \dots.
\]

\begin{example}[$\mathbb{C}$]
  \[
  \mathbb{C} = \mathbb{R}\left(i\right) = \mathbb{R}\left[i\right]
  \]
  $\mathbb{C}$ is also a \mynameref{def:vectorspace} generated by $1$
  and $i$: $\forall z \in \mathbb{C}$ it holds $z = x + i y$ where
  $x,y \in \mathbb{R}$.
\end{example}

\begin{proposition}
  The following assignment are equivalent
  \begin{enumerate}
  \item $\alpha$ is algebraic over $K$
  \item $K\left[\alpha\right]$ is a finite dimensional
    \mynameref{def:vectorspace} over $K$
  \item $K\left[\alpha\right] = K\left(\alpha\right)$
    \footnote{
      Contrary to the example \ref{ex:kalgebranotfield} we see
      that $K$-algebra is a field there.
    }
  \end{enumerate}
  \begin{proof}
    Lets proof that 1 implies 2. If $\alpha$ is algebraic over $K$
    then using lemma \mynameref{def:minpolynomial} $\exists
    P_{min}\left(\alpha, K\right)$:
    \[
    P_{min}\left(\alpha, K\right) = \alpha^d + a_{d-1}\alpha^{d-1} +
    a_1 \alpha + a_0 = 0,
    \]
    where $a_k \in K$. Then
    \[
    \alpha^d  = - a_{d-1}\alpha^{d-1} -
    a_1 \alpha - a_0
    \]
    this means that any $\alpha^n$ can be represented as a linear
    combination of finite number of powers of $\alpha$ i.e.
    $K\left[\alpha\right]$ generated by $1, \alpha, \dots,
    \alpha^{d-1}$ is a finite dimensional \mynameref{def:vectorspace}.

    Lets proof that 2 implies 3. Its enough to prove that
    $K\left[\alpha\right]$ is a field because
    $K\left[\alpha\right] \subset K\left(\alpha\right)$.
    
    Let $x \ne 0 \in
    K\left[\alpha\right]$ then lets look at an operation
    $x \cdot K\left[\alpha\right] \rightarrow
    K\left[\alpha\right]$. This is \mynameref{def:injection}.
    \footnote{
      If $y, z \in K\left[\alpha\right]$ and
      $\dim K\left[\alpha\right] = d < \infty$ where
      $d = \deg P_{min}\left(\alpha, K\right)$. Then
      $y = \sum_{i=0}^{d-1} y_i \alpha^i$ and
      $z = \sum_{i=0}^{d-1} z_i \alpha^i$ where $y_i, z_i \in K$.
      We have $y - z = \sum_{i=0}^{d-1} \left(y_i - z_i\right)
      \alpha^i \ne 0$ if $y \ne z$ (i.e.
      $\exists i: y_i \ne z_i$)
      because $y -z$ can be considered as a polynomial of degree
      $ \le d - 1 < \deg P_{min}\left(\alpha, K\right)$ and cannot be equal
      to 0 by minimal polynomial definition.
      Continue we have $x \cdot \left(y - z\right) \ne 0$ because it
      also can be considered as a product of 2 polynomial of degree
      $< d$. 
      Thus
      \[
      x \cdot y  \ne x \cdot z
      \]
      i.e. \mynameref{def:injection} property is satisfied.
    }
    But the $K\left[\alpha\right]$ is finite dimensional
    \mynameref{def:vectorspace} and a \mynameref{def:homomorphism} between
    2 vector spaces with the same dimension is
    \mynameref{def:surjection}
    \footnote{
      Two vector spaces with same dimension are isomorphic each other
      (see lemma \ref{lem:vsisomorphism})
    }
    thus $\exists y \in K\left[\alpha\right]$
    such that $x \cdot y = 1_{K\left[\alpha\right]}$. Therefore $x$ is
    invertable and $K\left[\alpha\right]$ is a
    \mynameref{def:field}.

    Lets proof that 3 implies 1. Let $K\left[\alpha\right]$ is a
    \mynameref{def:field} but $\alpha$ is not algebraic. Thus $\forall P
    \in K\left[X\right]$ $P(\alpha) \ne 0$. The we have an
    \mynameref{def:injection} \mynameref{def:homomorphism} $i$ :
    $K\left[X\right] \to K\left[\alpha\right] = K\left(\alpha\right)$
    which sends $P\left(X\right)$ to $P\left(\alpha\right)$.
    \footnote{
      And if $P\left(X\right) \ne 0$ then
      $P\left(\alpha\right) \ne 0$
    }
    But $K\left[X\right]$
    is not a field thus $K\left[\alpha\right]$ should not be a field
    too that is in contradiction with the initial conditions.
    \footnote{
      Alternative prove is the following.
      Let $x \ne 0 \in K\left[X\right]$ and $K\left[\alpha\right]$ is a
      field then $i(x)$ is invertable i.e. $\exists y \in
      K\left[X\right]:  i(x) i(y) = 1$ or 
      $i(xy) = 1$ or finally $x$ - is invertable and $K\left[X\right]$
      is a field but $K\left[X\right]$ is a ring and therefore we just
      got a contradiction. 
    }
  \end{proof}
  \label{prop:lec1_1}
\end{proposition}

\begin{definition}[Algebraic extension]
  $L$ an extension of $K$ is called algebraic over $K$ if $\forall
  \alpha \in L$ - $\alpha$ is algebraic over $K$. 
  \label{def:algebraicextension}
\end{definition}

\begin{proposition}
  If $L$ is algebraic over $K$ then any K-subalgebra of $L$ is a
  \mynameref{def:field}.
  \begin{proof}
    Let $L' \subset L$ is a subalgebra and let $\alpha \in L'$. We want
    to show that $\alpha$ is invertable. $\alpha$ is algebraic
    therefore $\alpha \in K\left[\alpha\right] \subset L' \subset L$
    and it's invertable.
    \footnote{
      As soon as $K\left[\alpha\right] = K\left(\alpha\right)$ is a
      field then its any element (especially $\alpha$) is invertable. 
    }
  \end{proof}
  \label{prop:lec1_algebraicsubalgebra}
\end{proposition}

\begin{proposition}
  Let $K \subset L \subset M$. $\alpha \in M$ - algebraic over $K$
  then $\alpha$ algebraic over $L$ and
  $P_{min}\left(\alpha, L\right)$ divides $P_{min}\left(\alpha,
  K\right)$. 
  \begin{proof}
    It is clear because $P_{min}\left(\alpha,K\right) \in
    L\left[X\right]$.
    \footnote{
      Thus $\exists P_L \in L\left[X\right]$ such that
      $P_L\left(\alpha\right) = 0$ i.e. $\alpha$ is algebraic over $L$.
      
      As soon as $P_{min}\left(\alpha,K\right) \in L\left[X\right]$ then
      using \mynameref{lem:minpolynomial} one can get that
      $P_{min}\left(\alpha,L\right)$ divides
      $P_{min}\left(\alpha,K\right)$.
    }
  \end{proof}
  \label{prop:lec1_algebraic}
\end{proposition}

We can consider the following example as an illustration for
proposition \ref{prop:lec1_algebraic}: 
\begin{myexample}
  $K = \mathbb{R}, L=M=\mathbb{C}$. $\alpha = i \in M$ is algebraic
  over $K = \mathbb{R}$ and therefore using the proposition
  \ref{prop:lec1_algebraic} it is
  algebraic over $L = \mathbb{C}$. Moreover $P_{min}(\alpha, L) = X
  -i$ and it divides $P_{min}(\alpha, K) = X^2 + 1$.
\end{myexample}

\section{Finite extensions. Algebraicity and finiteness}

\begin{definition}[Finite extension]
  $L$ is a finite extension of $K$ if $dim_k L < \infty$. $dim_k L$ is
  called as degree of $L$ over $K$ and is denoted by
  $\left[L:K\right]$
  \label{def:finiteextension}
\end{definition}

\begin{theorem}[The multiplicativity formula for degrees]
  Let $K \subset L \subset M$. Then $M$ is
  \mynameref{def:finiteextension} over $K$ if and only if
  $M$ is \mynameref{def:finiteextension} over $L$ and
  $L$ is \mynameref{def:finiteextension} over $K$. In this case
  \[
  \left[M:K\right] = \left[M:L\right] \left[L:K\right].
  \]
  \begin{proof}
    Let $\left[M:K\right] < \infty$ but any linear independent set of
    vectors  $\left\{m_1, m_2, \dots, m_n\right\}$ over $L$ is also
    linear independent over $K$ thus
    \[
    \left[M:K\right] < \infty \Rightarrow \left[M:L\right] < \infty
    \]
    also $L$ is a vector sub space of $M$ thus if
    $\left[M:K\right] < \infty$ then $\left[L:K\right] < \infty$.

    Let $\left[M:L\right] < \infty$ and $\left[L:K\right] < \infty$
    then we have the following basises
    \begin{itemize}
    \item $L$-basis over $M$: $\left(e_1, e_2, \dots, e_n\right)$
    \item $K$-basis over $L$:
      $\left(\varepsilon_1, \varepsilon_2, \dots, \varepsilon_d\right)$
    \end{itemize}
    Lets proof that $e_i\varepsilon_j$ forms a $K$-basis over $M$.
    $\forall x \in M$:
    \[
    x = \sum_{i=1}^n a_i e_i, 
    \]
    where $a_i \in L$ and can be also written as
    \[
    a_i = \sum_{j=1}^d b_{ij} \varepsilon_j,
    \]
    where $b_{ij} \in K$.
    Thus
    \[
    x = \sum_{i=1}^n \sum_{j=1}^d b_{ij} \varepsilon_j e_i, 
    \]
    therefore $\varepsilon_j e_i = e_i \varepsilon_j$ generates $M$
    over $K$. From the other side we should check that $\varepsilon_j
    e_i$ linear independent system of vectors. Lets
    \[
    \sum_{i,j} c_{ij} \varepsilon_j e_i =
    \sum_{i=1}^n \left( \sum_{j=1}^d c_{ij} \varepsilon_j \right) e_i = 0,
    \]
    then $\forall i$:
    \[
    \sum_{j=1}^d c_{ij} \varepsilon_j = 0.
    \]
    Thus $\forall i,j: c_{ik} = 0$ that finishes the proof the linear
    independence.   
    The number of linear independent vectors is $n \times d$ i.e.
    \[
    \left[M:K\right] = \left[M:L\right] \left[L:K\right].
    \]
  \end{proof}
  \label{thm:mulformuladegrees}
\end{theorem}

\begin{definition}[$K\left(\alpha_1, \dots, \alpha_n\right)$]
  $K\left(\alpha_1, \dots, \alpha_n\right) \subset L$ generated by
  $\alpha_1, \dots, \alpha_n$ is the smallest sub field of $L$
  contained  $K$ and $\alpha_i \in L$.
\end{definition}

\begin{theorem}[About towers]
  $L$ is finite over $K$ if and only if $L$ is generated by a finite
  number of algebraic elements over $K$.
  \begin{proof}
    If $L$ is finite then $\alpha_1, \dots, \alpha_d$ is a basis. In
    this case
    $L = K\left[\alpha_1, \dots, \alpha_d\right] = K\left(\alpha_1,
    \dots, \alpha_d\right)$. Moreover each
    $K\left[\alpha_i\right]$ is finite dimensional thus by
    proposition \ref{prop:lec1_1} $\alpha_i$ is algebraic.

    From other side if we have a finite set of algebraic elements
    $\alpha_1, \dots, \alpha_d$ then
    $K\left[\alpha_1\right]$ is a finite dimensional
    \mynameref{def:vectorspace} over $K$,
    $K\left[\alpha_1, \alpha_2\right]$ is a finite dimensional
    \mynameref{def:vectorspace} over $K\left[\alpha_1\right]$ and
    so on $K\left[\alpha_1, \dots, \alpha_d\right]$ is a finite dimensional
    \mynameref{def:vectorspace} over
    $K\left[\alpha_1, \dots, \alpha_{d-1}\right]$. All elements
    are algebraic thus
    \[
    K\left[\alpha_1, \dots, \alpha_i\right] =
    K\left(\alpha_1, \dots, \alpha_i\right)
    \]
    Then using theorem \ref{thm:mulformuladegrees} we can conclude
    that $K\left(\alpha_1, \dots, \alpha_d\right)$ has finite
    dimension. 
  \end{proof}
  \label{thm:lec1_2}
\end{theorem}

\section{Algebraicity in towers. An example}

\begin{theorem}
  $K \subset L \subset M$ then $M$ \mynameref{def:algebraicextension}
  over $K$ if and only if $M$ algebraic over $L$ and $L$ algebraic
  over $K$. 
  \begin{proof}
    If $\alpha \in M$ is an \mynameref{def:algebraicelement} over $K$ then
    $\exists P \in K\left[X\right]$ such that
    $P\left(\alpha\right) = 0$ but the
    polynomial $P \in K\left[X\right] \subset L\left[X\right]$
    thus $\alpha$ is algebraic over $L$.
    If $\alpha \in L \subset M$ then $\alpha$ is algebraic over $K$
    thus $L$ is algebraic over $K$.

    Let $M$ algebraic over $L$ and $L$ algebraic over $K$ and let
    $\alpha \in M$. We want to prove that $\alpha$ is algebraic over
    $K$. Lets consider $P_{min}\left(\alpha, L\right)$ the polynomial
    coefficients are from $L$ and they (as soon as they count is a
    finite)  generate a finite extension $E$ over $K$ thus
    $E\left(\alpha\right)$ is finite over $E$ (exists a relation
    between powers of $\alpha$) is finite over $K$ thus $\alpha$ is
    algebraic over $K$.
    \footnote{
      $P_{min}\left(\alpha, L\right) = \sum_{i = 0}^{d-1} l_i
      \alpha^i$ where $l_i \in L$ and each $l_i$ is algebraic over $K$
      by algebraic extension definition
      \ref{def:algebraicextension}. By theorem
      \ref{thm:lec1_2} there are finite number of $l_i$
      and they forms an algebraic extension
      $E = K\left(l_0, l_1, \dots, l_{d-1}\right)$. The
      $E\left(\alpha\right)$ is finite over $E$ and therefore finite
      over $K$. As soon as $E\left(\alpha\right)$ has a finite
      dimension over $K$ thus there exists a relation for powers of
      $\alpha$ such that $\sum_{i=0}^n k_i \alpha^i = 0$ i.e. 
      $\alpha$ is algebraic. 
    }
  \end{proof}
\end{theorem}
\begin{example}[$\mathbb{Q}$ extension]
  $\mathbb{Q}\left( \sqrt[3]{2}, \sqrt{3}\right)$ algebraic and finite
  over $\mathbb{Q}$:
  \[
  \mathbb{Q} \subset \mathbb{Q}\left( \sqrt[3]{2}\right)
  \subset \mathbb{Q}\left( \sqrt[3]{2}, \sqrt{3}\right)
  \]

  Minimal polynomial
  \[
  P_{min}\left(\sqrt[3]{2}, \mathbb{Q}\right) = x^3 - 2.
  \]

  $\mathbb{Q}\left( \sqrt[3]{2}\right)$ is generated over $\mathbb{Q}$
  by $1, \sqrt[3]{2}, \sqrt[3]{4}$ thus
  $\left[\mathbb{Q}\left( \sqrt[3]{2}\right): \mathbb{Q}\right] = 3$.

  But $\sqrt{3} \notin \mathbb{Q}\left( \sqrt[3]{2}\right)$ because
  otherwise $\left[\mathbb{Q}\left( \sqrt{3}\right): \mathbb{Q}\right]
  = 2$ must devide  
  $\left[\mathbb{Q}\left( \sqrt[3]{2}\right): \mathbb{Q}\right] = 3$
  that is impossible.

  Therefore $x^2 - 3$ is irreducible over
  $\mathbb{Q}\left( \sqrt[3]{2}\right)$ and
  \[
  P_{min}\left(\sqrt{3}, \mathbb{Q}\left( \sqrt[3]{2}\right)\right) =
  x^2 - 3.
  \]

  \[
  \left[\mathbb{Q}\left( \sqrt[3]{2}, \sqrt{3}\right):
    \mathbb{Q}\right] = 3 \cdot 2 = 6.
  \]
\end{example}

\begin{proposition}[On dimension of extension]
  \[
  \left[K\left(\alpha\right) : K\right] =
  \deg\left(P_{min}\left(\alpha, K\right)\right),
  \]
  if $\alpha$ is algebraic.
  \begin{proof}
    If $\deg\left(P_{min}\left(\alpha, K\right)\right) = d$ then $1, \alpha,
    \cdots, \alpha^{d-1}$ - $d$ independent vectors and dimension
    $K\left(\alpha\right)$ is $d$.
  \end{proof}
  \label{prop:dimextension}
\end{proposition}

\begin{proposition}[About algebraic closure]
  If $K \subset L$ ($L$ extension of $K$). Consider
  \[
  L' = \left\{
  \alpha \in L \mid \alpha \mbox{ algebraic over } K
  \right\},
  \]
  then $L'$ sub-field of $L$ and is called as algebraic closure of $K$
  in $L$.
  \begin{proof}
    We have to prove that if $\alpha, \beta$ are algebraic then
    $\alpha + \beta$ and $\alpha \cdot \beta$ are also algebraic. This
    is trivial  because
    \[
    \alpha + \beta, \alpha \cdot \beta \in K\left[\alpha, \beta\right]
    \]
    \footnote{
      We also have that $K\left[\alpha, \beta\right]$ is a field:
      $K\left[\alpha, \beta\right]
      = K\left(\alpha, \beta\right)$.
      Really
      $K\left[\alpha\right] =
      K\left(\alpha\right)$ (see proposition
      \ref{prop:lec1_1}). $\beta$ is algebraic over $K$ and 
      therefore over $K\left(\alpha\right)$ thus we can construct
      $K\left(\alpha\right)\left[\beta\right] = K\left(\alpha, \beta\right)$
      by proposition \ref{prop:lec1_1}
    }
  \end{proof}
\end{proposition}


\section{A digression: Gauss lemma, Eisenstein criterion}

What we have seen so far:

\begin{itemize}
\item $K$ is a field, $\alpha$ is an \mynameref{def:algebraicelement}
  over $K$ if it is a root of a polynomial $P \in K\left[X\right]$.
\item $L$ is an \mynameref{def:algebraicextension} over $K$ if
  $\forall \alpha \in L$: $\alpha$ is an algebraic over $K$
\item $L$ is a \mynameref{def:finiteextension} over $K$ if $dim_K L <
  \infty$.
\item If an extension is finite then it is algebraic
\item An extension is finite if and only if it is algebraic and
  generated by a finite number of algebraic elements (see theorem
  \ref{thm:lec1_2})
\item $\left[K\left[\alpha\right]:K\right] =
  deg P_{min}\left(\alpha, K\right)$ (see proposition
  \ref{prop:dimextension}). 
\end{itemize}

How to decide that a polynomial $P$ is irreducible over $K$?
About polynomial $x^3 - 2$ it is easy to decide that it's irreducible
over $\mathbb{Q}$, but what's about $x^{100}-2$?

\begin{lemma}[Gauss]
  Let $P \in \mathbb{Z}\left[X\right]$, i.e. a polynomial with integer
  coefficients, then if $P$ decomposes over $\mathbb{Q}$ ($P = Q\cdot
  R, deg Q,R < deg P$) then it also decomposes over $\mathbb{Z}$.
  \begin{proof}
    Let $P = Q R$ over $\mathbb{Q}$. Then
    \begin{eqnarray}
      m Q = Q_1, Q_1 \in \mathbb{Z}\left[X\right],
      \nonumber \\
      n R = R_1, R_1 \in \mathbb{Z}\left[X\right],
      \nonumber
    \end{eqnarray}
    thus
    \[
    n m P = Q_1 R_1. 
    \]
    There exists $p$ that divides $mn$: $p \mid mn$ thus in modulo $p$
    we have
    \[
    0 = \overline{Q_1}\overline{R_1}
    \]
    but $p$ is prime and the equation is in the field $\mathbb{F}_p$
    thus either $\overline{Q_1} = 0$ or $\overline{R_1} = 0$. Let
    $\overline{Q_1} = 0$ thus $p$ divides all coefficients in $Q_1$
    and we can take $\frac{Q_1}{p} = Q_2 \in
    \mathbb{Z}\left[X\right]$. Continue for all primes in $mn$ we can
    get that
    \[
    P = Q_s R_t,
    \]
    where $Q_s, R_t \in \mathbb{Z}\left[X\right]$.
  \end{proof}
  \label{lem:gauss}
\end{lemma}

\begin{example}[Eisenstein criterion]
  Lets consider the following polynomial $x^{100} -2$. It's
  irreducible. Lets prove it. If it reducible then
  $\exists Q, R \in \mathbb{Z}\left[X\right]$ such that
  \begin{equation}
    x^{100} -2 = Q R
    \label{eq:ex_eisenstein}
  \end{equation}
  Lets consider (\ref{eq:ex_eisenstein}) modulo 2. In the case we will
  have
  \[
  Q R \equiv x^{100} \mod 2,
  \]
  therefore
  \begin{eqnarray}
    Q \equiv x^k \mod 2,
    \nonumber \\
    R \equiv x^l \mod 2,
    \nonumber
  \end{eqnarray}
  or
  \[
  Q = x^k + \dots + 2 \cdot m
  \]
  and
  \[
  R = x^l + \dots + 2 \cdot n
  \]
  thus
  \[
  QR = x^{100} + 4 \cdot nm
  \]
  that is impossible because $n,m \in \mathbb{Z}$ and $nm \ne
  -\frac{1}{2}$. 
  \label{ex:eisenstein}
\end{example}

\begin{lemma}[Eisenstein criterion]
  Lets $P \in \mathbb{Z}\left[X\right]$ and
  $P = a_n X^n + a_{n-1} X^{n-1} + a_1 X + a_0$. If $\exists p$ -
  prime such that $p \nmid a_n$, $p \mid a_i \forall i < n$ and
  $p^2 \nmid a_0$, then $P \in \mathbb{Z}\left[X\right]$ is
  irreducible. 
  \begin{proof}
    the same as for example \ref{ex:eisenstein}.
  \end{proof}
  \label{lem:eisenstein}
\end{lemma}

Note: that both: \mynameref{lem:gauss} and \mynameref{lem:eisenstein} are
valid by replacing $\mathbb{Z}$ with an \mynameref{def:ufd} $R$ and
$\mathbb{Q}$ by its factorization field.